How to Calculate the Cross Product of Two Vectors

Consider two general three-dimensional vectors defined in Cartesian coordinates., Set up the matrix., Calculate the determinant of the matrix.

3 Steps 1 min read Easy

Step-by-Step Guide

  1. Step 1: Consider two general three-dimensional vectors defined in Cartesian coordinates.

    a=Ai+Bj+Ckb=Di+Ej+Fk{\displaystyle {\begin{aligned}{\mathbf {a} }&=A{\mathbf {i} }+B{\mathbf {j} }+C{\mathbf {k} }\\{\mathbf {b} }&=D{\mathbf {i} }+E{\mathbf {j} }+F{\mathbf {k} }\end{aligned}}} Here, i,j,k{\displaystyle {\mathbf {i} },{\mathbf {j} },{\mathbf {k} }} are unit vectors, and A,B,C,D,E,F{\displaystyle A,B,C,D,E,F} are constants. , One of the easiest ways to compute a cross product is to set up the unit vectors with the two vectors in a matrix. a×b=|ijkABCDEF|{\displaystyle {\mathbf {a} }\times {\mathbf {b} }={\begin{vmatrix}{\mathbf {i} }&{\mathbf {j} }&{\mathbf {k} }\\A&B&C\\D&E&F\end{vmatrix}}} , Below, we use cofactor expansion (expansion by minors). a×b=(BF−EC)i−(AF−DC)j+(AE−DB)k{\displaystyle {\mathbf {a} }\times {\mathbf {b} }=(BF-EC){\mathbf {i} }-(AF-DC){\mathbf {j} }+(AE-DB){\mathbf {k} }} This vector is orthogonal to both a{\displaystyle {\mathbf {a} }} and b.{\displaystyle {\mathbf {b} }.}
  2. Step 2: Set up the matrix.

  3. Step 3: Calculate the determinant of the matrix.

Detailed Guide

a=Ai+Bj+Ckb=Di+Ej+Fk{\displaystyle {\begin{aligned}{\mathbf {a} }&=A{\mathbf {i} }+B{\mathbf {j} }+C{\mathbf {k} }\\{\mathbf {b} }&=D{\mathbf {i} }+E{\mathbf {j} }+F{\mathbf {k} }\end{aligned}}} Here, i,j,k{\displaystyle {\mathbf {i} },{\mathbf {j} },{\mathbf {k} }} are unit vectors, and A,B,C,D,E,F{\displaystyle A,B,C,D,E,F} are constants. , One of the easiest ways to compute a cross product is to set up the unit vectors with the two vectors in a matrix. a×b=|ijkABCDEF|{\displaystyle {\mathbf {a} }\times {\mathbf {b} }={\begin{vmatrix}{\mathbf {i} }&{\mathbf {j} }&{\mathbf {k} }\\A&B&C\\D&E&F\end{vmatrix}}} , Below, we use cofactor expansion (expansion by minors). a×b=(BF−EC)i−(AF−DC)j+(AE−DB)k{\displaystyle {\mathbf {a} }\times {\mathbf {b} }=(BF-EC){\mathbf {i} }-(AF-DC){\mathbf {j} }+(AE-DB){\mathbf {k} }} This vector is orthogonal to both a{\displaystyle {\mathbf {a} }} and b.{\displaystyle {\mathbf {b} }.}

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Kimberly Ramos

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